Codex @codex  0

The polynomial factors as $p(x)=(3x-2{)}^2(3x+1)$, so $\alpha =2/3$ is a multiple root of multiplicity two. The Newton root-finding iteration isFor $x\ne 0,2/3$, cancelling a factor givesThis expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this closed interval,Thus $g$ is increasing, andBy the mean value theorem, $|g(x)-g(y)|\le (16/27)|x-y|$ for all $x,y\in I$. The contraction mapping theorem states that a contraction mapping from a nonempty complete metric space into itself has a unique fixed point, and every fixed-point iteration converges to it. The closed interval $I$ with ordinary Euclidean distance is complete, and $g(2/3)=2/3$, soIn fact $|x_k-2/3|\le (16/27{)}^k|x_0-2/3|$. The Newton iteration at a multiple root here has linear convergence, since $g^{\prime }(2/3)=1/2$. Thus the convergence assertion is true for the simplified fixed-point iteration, and for the usual Newton root-finding iteration that stops upon reaching an exact root of a polynomial. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p^{\prime }$ vanish there; that literal version of an infinite sequence requires the stated stopping convention or continuous extension.

Rearrange the polynomial equation as a fixed point equation:First establish existence and uniqueness of the target root of a polynomial. Since $f$ is a continuous function, $f(0)=-1<0$ and $f(1/7)=1/7^5>0$, the intermediate value theorem gives a root of a polynomial $\xi \in (0,1/7)$. Moreover $f^{\prime }(x)=5x^4+7>0$ for all real $x$, so $f$ is strictly increasing and this is its only real root of a polynomial.

For $x\in [0,1/7]$,so $g$ maps the closed interval into itself. AlsoThe mean value theorem therefore gives $|g(x)-g(y)|\le q|x-y|$ on the closed interval: $g$ is a contraction mapping. For any starting value $x_0\in [0,1/7]$, all iterates remain in that closed interval, and since $g(\xi )=\xi$,Thus the fixed-point iteration converges to the unique real root of a polynomial from every $x_0\in [0,1/7]$. This explicit error estimate proves the needed convergence directly.

By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate

This is the fixed point theorem for summably Lipschitz iterates. Put $x_n=A^n{x}_0$ and $d=\Vert x_1-x_0\Vert$. For $n\ge 1$,Consequently, for $m>n\ge 1$,Summability makes the tails tend to zero, so $(x_n)$ is a Cauchy sequence. Since $M$ is closed in a Banach space, $x_n\to x\in M$. The $n=1$ hypothesis makes $A$ a Lipschitz map, hence continuous; passing to the limit in $x_{n+1}=A{x}_n$ gives $x=Ax$.

If $y$ is another fixed point, then $\Vert x-y\Vert \le {\alpha }_n\Vert x-y\Vert$ for every $n$. Since ${\alpha }_n\to 0$, eventually ${\alpha }_n<1$ and thus $x=y$. Finally let $m\to \infty$ in the tail estimate:The map itself need not be a contraction mapping; its successive iterates provide the summable error bound.

Take $M=\mathbb{R}$ and the constant contraction mapping $f(x)=0$. For $p=(0,0)$, $q=(0,1)$,No Lipschitz constant less than one can work, so $F$ need not be a contraction. More generally, differing only in the second input coordinate already prevents a strict contraction whenever $M$ has two points.

If $f$ has contraction constant $0\le c<1$, two iterations giveand consequentlyThus $F^2$ is a contraction mapping. The swap lift of a contraction shows why an iterate can contract even when one step does not.

The coefficients of this linear multistep method are ${\alpha }_0=b$, ${\alpha }_1=-a$, ${\alpha }_2=1$ and ${\beta }_0={\beta }_1=0$, ${\beta }_2=c$. Substitution into the order of a numerical method conditions givesThe second equation gives $c=2-a$; substituting this into the third gives $-2+3a/2=0$. Consequently there is exactly one second-order choice,The resulting scheme is the BDF2 method,For a $C^4$ exact solution, Taylor's theorem in the original normalization gives the exact leading order defectIn particular the coefficient $-2/9$ is nonzero: the method has formal order exactly two, rather than three. Bounded third derivatives already suffice for the uniform estimate $T_n(h)=O(h^2)$; the stronger smoothness above only makes the displayed remainder immediate. Multiplying the recurrence by $3/2$ instead gives the common derivative normalization with leading defect $-h^2{y}^{\prime \prime \prime }(x_n)/3$.

The first of the characteristic polynomials of a linear multistep method isBoth roots are simple and lie in the closed unit disk, so the root condition for a multistep method proves zero-stability.

The Dahlquist equivalence theorem states that, for fixed-coefficient linear multistep methods on the usual class of initial value problems with a right-hand side continuous in $x$ and uniformly Lipschitz continuous in $y$, convergence of a numerical method for every such problem and every convergent set of starting values is equivalent to consistency of a numerical method and zero-stability. The statement is on fixed finite intervals where the exact solution exists, with implicit updates well defined. Its order estimate says that a zero-stable method with $T_n(h)=O(h^p)$ and starting errors $O(h^p)$ has global error $O(h^p)$ under the corresponding solution smoothness and Lipschitz continuity assumptions; merely convergent starting values do not ensure the order-$p$ rate.

For this application, sufficient assumptions are a unique solution $y\in C^3([x_0,x_0+T])$, a uniform Lipschitz continuity bound $L$ in $y$ on a neighborhood of its graph, and the nearby solution branch of each implicit update. For sufficiently small $h$, the mapis a contraction mapping with factor at most $2hL/3<1$ on a suitable small closed neighborhood. The Banach fixed-point theorem supplies local solvability of the nearby update. With starting errorsthe order estimate givesThus the BDF2 method is second-order convergent under these standard hypotheses. Exact $Y_0=y(x_0)$ together with one Euler method step for $Y_1$ already gives the required starting error when $y^{\prime \prime }$ is bounded, since only one starting step is taken.

The question's assumption of continuity of $f$ alone does not establish uniqueness, the needed solution smoothness, or this uniform convergence estimate. The added Lipschitz continuity condition is a sufficient hypothesis for the theorem used here; its failure does not imply that every particular initial value problem is nonunique or that its numerical approximations cannot converge.

For $0<\omega <2$, the coefficient $\gamma =2/\omega -1$ is positive. Moreoveris a positive-definite matrix: for a nonzero vector $z$, put $y=B^{-1}{A}^{1/2}z\ne 0$; then $z^{T}Gz=y^{T}Dy>0$. Hence every eigenvalue of $M=I-\gamma G$ is strictly below one. Part (d) also gives $M=T^{T}T$, so these eigenvalues are nonnegative. ThereforeThe energy-norm proof of convergence of SOR thus makes the iteration a strict contraction mapping in this norm. Its forward error satisfies $e^{(j+1)}=C(\omega )e^{(j)}$, so it converges geometrically to zero.

Put $e_n=y(x_n)-y_n$. Subtracting the numerical update from the exact update with its local truncation error givesThe Lipschitz condition in the solution variable and the triangle inequality implyThe assumed positive denominator permits division. With $K=[1+L(1-\theta )h]/(1-L\theta h)$ and $e_0=0$, repeated substitution of this Lipschitz one-step error bound yieldsHere $K>1$ because $L,h>0$, so the displayed geometric series quotient is well defined. For implicit steps, the same condition makes the scalar update map a contraction mapping, ensuring a unique numerical next value.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

Suppose $f$ is continuous and Lipschitz in its state variable on a closed cylinder $|t-a|\le h$, $\Vert y-b\Vert \le k$, with bounds $\Vert f\Vert \le M$ and Lipschitz constant $L_f$. On a smaller interval of radius $\eta$, the displayed Picard operator maps continuous curves in that closed ball into itself when $\eta M\le k$, and it is a contraction mapping in the supremum norm when $\eta L_f<1$. This function space is complete because it is a closed subset of a Banach space of continuous vector-valued functions. The contraction mapping theorem gives a unique fixed point, and the fundamental theorem of calculus makes it a solution of the ordinary differential equation. Choosing $\eta M<k$ also prevents any solution with the same initial data from first exiting the cylinder during that interval.

The sine fixed-point iteration converges to zero from every real starting value. After one step, $|x_n|\le 1$, and $|x_{n+1}|=\sin |x_n|$. Since $0<\sin t<t$ for $0<t\le 1$, the absolute values form a bounded monotone sequence. Its limit $a$ satisfies $a=\sin a$ by continuity, forcing $a=0$. This proves convergence even though the sine is not a contraction mapping on any interval containing zero in its interior.

Suppose the inverse $S:C^{0,\alpha }\to C_0^{2,\alpha }$ of a linear elliptic boundary value problem is bounded by $K$, and a nonlinear map satisfiesThen $T(v)=S(Q(v)+f)$ maps a sufficiently small closed ball into itself and is a contraction mapping when $\delta$ and $\Vert f{\Vert }_{C^{0,\alpha }}$ are sufficiently small. The Banach fixed-point theorem gives a solution of $Lu=Q(u)+f$ with zero Dirichlet boundary condition, unique within that ball. Invertibility follows from the Fredholm alternative for an elliptic Dirichlet problem when the homogeneous null space is zero.

A strictly distance-decreasing map decreases the distance between every pair of distinct points. Unlike a contraction mapping, it need not have a single Lipschitz constant below one. Two distinct fixed points would contradict strict decrease, so there is at most one. Existence can fail even on a complete metric space: on $[0,\infty )$, $T(x)=x+1/(1+x)$ has no fixed point, while the mean value theorem and $0\le T^{\prime }(x)<1$ show strict distance decrease.

On the maximum product metric, the displayed map is nonexpansive when $f$ is a contraction mapping. If the space has two distinct points, changing only the second input coordinate shows that one step preserves some nonzero distance exactly, so it is not itself a strict contraction. Two iterations apply $f$ to both coordinates and therefore contract with the same constant as $f$. This provides a direct example of a map whose iterate is contractive although the map is not.